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State set of values for which K has 3 solutions for eq: 2 + 9x + 3x^2 - x^3 - k = 0

Not really sure how I should approach this question...
Original post by TSR360
Not really sure how I should approach this question...


"State" usually implies minimal, or no, working, which makes me think this is the latter part of a multi-part question. Please post the entire question and any working/results so far.
Reply 2
Original post by ghostwalker
"State" usually implies minimal, or no, working, which makes me think this is the latter part of a multi-part question. Please post the entire question and any working/results so far.

I can't. This is a really old question I found when looking through my notes & noticed the w/o didn't make any sense. Here is what it looks like in my notes: http://prntscr.com/tgpmxf
(edited 3 years ago)
Original post by TSR360
I can't. This is a really old question I found when looking through my notes & noticed the w/o didn't make any sense. Here is what it looks like in my notes: http://prntscr.com/tgpmxf



OK.

If you look at the graph, I think it should be clear that there will be three solutions as long as k (defining the line y=k) is less than the maximum stationary point and greater than the minimum stationary point. I.e. the line y=k will cut the cubic in three places.

Hence work out the stationary points, and the rest follows.
Reply 4
Original post by TSR360
I can't. This is a really old question I found when looking through my notes & noticed the w/o didn't make any sense. Here is what it looks like in my notes: http://prntscr.com/tgpmxf

OK those notes are a bit mangled - dk/dx doesn't really make any sense - but if you consider your original cubic as a function of x you can ask where it's stationary points are. The working out shows that there is a min and a max at 2 real points, so by knowing the shape of such a cubic graph you can image translating it vertically by various amounts (positive and negative) to see how this affects the number of axis crossing points.

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